Precalculus (6th Edition)

Published by Pearson
ISBN 10: 013421742X
ISBN 13: 978-0-13421-742-0

Chapter 11 - Further Topics in Algebra - 11.2 Arithmetic Sequences and Series - 11.2 Exercises - Page 1024: 23


$a_n=3+2n$ $a_8=19$

Work Step by Step

$5, 7, 9, ...$ We have $a_1=5$ and $d=7-5=2$. Using the formula $a_n=a_1+(n-1)d$, we get: $a_n=5+(n-1)*2$ $a_n=5+(2n-2)$ $\boxed{a_n=3+2n}$ $a_8=3+2*8=3+16=\boxed{19}$.
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